How do you differentiate #y=(x^3-x^2-3)/(x^5+3)# using the quotient rule?

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Answer 1 · 2017-03-30T06:13:41

#(dy)/(dx)=(x^4+15x^2-4x)/(x^2+5)^2#

Quotient rule states if #y=y(x)=(g(x))/(h(x))#

then #(dy)/(dx)=((dg)/(dx)xxh(x)-(dh)/(dx)xxg(x))/(h(x))^2#

Here #g(x)=x^3-x^2-3# hence #(dg)/(dx)=3x^2-2x#

and as #h(x)=x^2+5#, #(dh)/(dx)=2x#

Hence #(dy)/(dx)=((3x^2-2x)xx(x^2+5)-2x xx (x^3-x^2-3))/(x^2+5)^2#

= #(3x^4-2x^3+15x^2-10x-2x^4+2x^3+6x)/(x^2+5)^2#

= #(x^4+15x^2-4x)/(x^2+5)^2#